Answer to Question #87048 in Mechanics | Relativity for Biraj

Question

A mass m, free to move along a line is attracted towards a given point on the line with a force proportional to its distance from the given point. If the mass starts from rest at a distance Xo from the given point, show that the mass moves in a simple harmonic motion.

Answer 1

Let P - given point.

Coordinate x of P = 0.

Coordinate x of m = x.

Then $F = -kx$ , so

$ma = -kx \\ ma + kx = 0 \\ m\ddot{x} + kx = 0$

We have linear differential equation,

Characteristic equation:

$m{\alpha}^2 + k = 0 \\ \alpha = \pm i \cdot \omega, \omega = \sqrt{\frac{k}{m}}$

So

$x = C_1 \cdot e^{t\alpha_1} + C_2 \cdot e^{t\alpha_2} = C_1(e^{it\omega})+C_2(e^{-it\omega})$

with Euler's formula:

$x = C_1(\cos(\omega t) + i\sin(\omega t)) + C_2(\cos(-\omega t) + i\sin(-\omega t))$

$x = (C_1 + C_2)cos(\omega t)) + (iC_1 - iC_2)(sin(\omega t))$

Let:

$C_1 + C_2 = C_3 \\ iC_1 + iC_2 = C_4 \\ \sqrt{C_3^2+C_4^2} = A \\$

Then:

$(\frac{C_3}{A})^2 + (\frac{C_4}{A})^2 = 1 \\ \frac{C_3}{A} = \sin \phi \\ \frac{C_4}{A} = \cos \phi$

$x = C_3\cos(\omega t) + C_4\sin(\omega t)$

$x = A(\sin \phi \cos(\omega t) + \cos \phi \sin(\omega t))$

$$x = A\sin(\omega t + \phi)$$

- harmonic motion equation.